I am self-learning Real Analysis. I'd like for someone to verify, if my exposition and proofs are sufficiently rigorous and technically correct. Thanks so much!
[K&N 3.1.2, Abbot 6.2.9] Assume that $\{f_n\}$ and $\{g_n\}$ are uniformly convergent sequences of functions.
(a) Show that $\{f_n + g_n\}$ is a uniformly convergent sequence of functions.
Proof. (My Attempt)
There exists $N_1$, such that $|f_n(x) - f(x)|<\epsilon/2$ for all $x \in A$ and for all $n \geq N_1$. There exists $N_2$, such that $|g_n(x) - g(x)| < \epsilon/2$ for all $x \in A$ and for all $n \geq N_2$.
Pick $N = \max{N_1,N_2}$.
\begin{align*}|f_n(x) + g_n(x) - (f(x) + g(x))| &= |f_n(x) - f(x) + g_n(x) - g(x)|\\&\leq |f_n(x) - f(x)| + |g_n(x) - g(x)| \quad \{ \text{Triangle Inequality} \} \\&< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\end{align*}
for all $n \geq N$ and $x \in A$. So, $(f_n + g_n)$ converges uniformly to $f + g$.
(b) Give an example to show that the product $(f_n g_n)$ may not converge uniformly.
Proof.
Define $f_n(x) = \frac{1}{n}$ and let $f(x) = 0$. Then, if $N > \frac{1}{\epsilon}$, we have $|f_n(x) - f(x)|<\epsilon$, for all $n \geq N$ and for all $x \in [0,\infty)$. Consequently, $(f_n)$ converges uniformly to $f$.
Define $g_n(x) = x$ and let $g(x) = x$. Then, $(g_n)$ converges uniformly to $g$.
$(f_n \cdot g_n)(x) = \frac{x}{n}$ and $(fg)(x) = 0$. However, $(f_n g_n)$ does not converge uniformly to $fg$ on $[0,\infty)$.
For example, let $\epsilon_0 = \frac{1}{2}$.
If $N = 1$, then for $x > \frac{1}{2}$, $|f_1g_1(x) - fg(x)| \geq \epsilon_0$.
If $N=2$, then for $x > 1, |f_2g_2(x) - fg(x)| \geq \epsilon_0$.
If $N=3$, then for $x > \frac{3}{2}, |f_3g_3(x) - fg(x)| \geq \epsilon_0$.
For any $N$, if $x > \frac{N}{2}, |(f_N \cdot g_N)(x) - fg(x)| \geq \epsilon_0$.
(c) Prove that, if there exists an $M > 0$ such that $|f_n| \leq M$ and $|g_n \leq M|$ for all $n \in \mathbf{N}$, then $(f_ng_n)$ converges uniformly.
Proof.
We are interested to prove that, when all the terms in the sequence of functions, $(f_n)$ and $(g_n)$, are bounded, then the product of sequences converges uniformly to the product of the respective limits.
\begin{align*}|f_n(x)g_n(x) - f(x)g(x)| &= |f_n(x)g_n(x) - f(x)g_n(x) + f(x)g_n(x) - f(x)g(x)| \\ &\leq |g_n(x)||f_n(x) - f(x)| + |f(x)||g_n(x) - g(x)|\end{align*}
Now, $|g_n(x)| \leq M$ for all $n \in \mathbf{N}$.
Moreover, $|f_n(x)|\leq M$ for all $n \in N$, for all $x \in A$. Consequently, $\lim_{n \to \infty} |f_n(x)| = |f(x)| \leq M$ for all $x \in A$, by the order limit theorem for sequences of real numbers.
There exists $N_1$, such that $|f_n(x) - f(x)| < \frac{\epsilon}{2M}$ for all $n \geq N_1$ and for all $x \in A$.
There exists $N_2$, such that $|g_n(x) - g(x)| < \frac{\epsilon}{2M}$ for all $n \geq N_2$ and for all $x \in A$.
Pick $N = \max \{ N_1,N_2\}$. Then,
\begin{align*}|f_n(x) g_n(x) - f(x) g(x)| &\leq |g_n(x)||f_n(x) - f(x)| + |f(x)||g_n(x) - g(x)|\\&< M\cdot \frac{\epsilon}{2M} + M \cdot \frac{\epsilon}{2M}\\&= \epsilon\end{align*}
for all $n \geq N$ and $x \in A$.
This closes the proof.